Answer:
(a) 2.34 s
(b) 6.71 m
(c) 38.35 m
(d) 20 m/s
Explanation:
u = 20 m/s, theta = 35 degree
(a) The formula for the time of flight is given by


T = 2.34 second
(b) The formula for the maximum height is given by


H = 6.71 m
(c) The formula for the range is given by


R = 38.35 m
(d) It hits with the same speed at the initial speed.
Answer:
a = 120 m/s²
Explanation:
We apply Newton's second law in the x direction:
∑Fₓ = m*a Formula (1)
Known data
Where:
∑Fₓ: Algebraic sum of forces in the x direction
F: Force in Newtons (N)
m: mass (kg)
a: acceleration of the block (m/s²)
F = 1200N
m = 10 kg
Problem development
We replace the known data in formula (1)
1200 = 10*a
a = 1200/10
a = 120 m/s²
Answer:
6 m/s
Explanation:
12m / 2s = 6 m/s
Hope that's the answer you seek.
Answer:
F = 69.5 [N]
Explanation:
We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.

where:
N = normal force [N]
miu = friction coefficient
f = friction force = 22 [N]
Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

where:
F = force exerted [N]
f = friction force [N]
m = mass = 95 [kg]
a = acceleration = 0.5 [m/s²]
Now replacing:
![F - 22 = 95*0.5\\F = 47.5 + 22\\F = 69.5 [N]](https://tex.z-dn.net/?f=F%20-%2022%20%3D%2095%2A0.5%5C%5CF%20%3D%2047.5%20%2B%2022%5C%5CF%20%3D%2069.5%20%5BN%5D)
Answer:
The object will travel 675 m during that time.
Explanation:
A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.
In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.
In this case, the position is calculated using the expression:
x = xo + vo*t + ½*a*t²
where:
- x0 is the initial position.
- v0 is the initial velocity.
- a is the acceleration.
- t is the time interval in which the motion is studied.
In this case:
- x0= 0
- v0= 0 because the object is initially stationary
- a= 6

- t= 15 s
Replacing:
x= 0 + 0*15 s + ½*6
*(15s)²
Solving:
x=½*6
*(15s)²
x=½*6
*225 s²
x= 675 m
<u><em>
The object will travel 675 m during that time.</em></u>